Inequality.$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$ 
Let $a,b,c \gt 0$. Prove that (Using Cauchy-Schwarz) :
  $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$$

I tried to use Cauchy-Schwarz in the following form
$$\sqrt{Ax}+\sqrt{By}+\sqrt{Cz}\leq \sqrt{(A+B+C)(x+y+z)}\tag{1}.$$ 
I wrote $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}=\frac{\sqrt{2a(c+a)(a+b)}+\sqrt{2b(b+c)(a+b)}+\sqrt{2c(b+c)(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}.$$
and then I applied on $(1)$: 
\begin{eqnarray}
A &=& 2a(c+a)   &\mbox{and}&  x=a+b;\\ 
B &=& 2b(a+b)   &\mbox{and}&  y=b+c;\\
C &=& 2c(b+c)   &\mbox{and}&  z=c+a ,
\end{eqnarray}
but I did not obtain anything. Thanks for your help. :)
 A: Denote $$S = \sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}}$$
The Inequality doesn't hold. Clearly, taking $b=c=0.05, a=5$ implies $$\sqrt{2}S>\sqrt{\frac{2a}{b+c}} = 10 > 3$$ By this method, it is easily shown that no upper bound exists.
For a lower bound, note firstly that when $a = b, c \rightarrow 0$, $S \rightarrow 2$. We'll prove that $S \ge 2$ for non negative reals $a,b,c$. 
We have, $$\frac{a+b+c}{2a} = \frac 12 \left(\frac{b+c}{a} + 1 \right) \ge \sqrt{\frac{b+c}{a}} \\ \implies \sqrt{\frac{a}{b+c}} \ge \frac{2a}{a+b+c}$$ Adding up the other two similar inequalities, we get the result.
A: You got it wrong. It should be:
$$
\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}} \leq 3
$$
See my comment on Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$ for a proof.
A: We will begin with an observation. Consider the function $$f(a,b,c)=\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} .$$ It has the property that $$f(ta,tb,tc)=f(a,b,c).$$ We may thus assume without loss of generality that $$a+b+c=1$$, by setting $$t=\frac{1}{a+b+c}.$$ Under this assumption, we may rewrite the function as $$\sqrt{\frac{2a}{1-a}}+\sqrt{\frac{2b}{1-b}}+\sqrt{\frac{2c}{1-c}}$$ If $a=1-\epsilon$ for $\epsilon>0$, $b+c=\epsilon$. Now note that in order for these square roots to be defined over the reals, then $0\leq a,b,c<1$, which means that $b,c<\epsilon$. Now for such an $\epsilon$,  $$\sqrt{\frac{2b}{1-b}}, \sqrt{\frac{2c}{1-c}}$$ are both close to zero, and $$\sqrt{\frac{2a}{1-a}}$$ is large.
A: By AM-HM, $\frac{2(a+b+c)}{3}\ge\frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}$$\implies(2a+2b+2c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge3^2$.
By Cauchy-Schwarz, $(2a+2b+2c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\left(\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{a+c}}\right)^2$.
Thus, $\left(\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{a+c}}\right)^2\le3^2$$\implies\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{a+c}}\le3$.
